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I am looking at the following image (from http://www.schoolphysics.co.uk/age16-19/Atomic%20physics/X%20rays/text/X_ray_spectra/index.html):

enter image description here

and I want to know, of the two curves, which one represents the element with the higher atomic number. That is, I understand that X-rays are scattered and that the peaks are characteristic of a material. What I am less clear on i why one curve should be above another relative to their atomic numbers (I am going to assume the cutoff voltage on the left is identical even though in this picture it isn't).

The reason I am confused is that I am told that that $\frac{1}{\sqrt{\lambda}} \propto Z$ which is all very well, but on this graph it would seem to indicate that the upper line is a heavier (higher Z) material. Is hat the case, or is intensity actually proportional to $\frac{1}{\sqrt{\lambda}}$ as well? I would have thought the lower curve would be a higher Z since it usually take less energy to knock electrons off of elements further up the periodic table.

Anyhow, what clarification might be offered is appreciated.

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    $\begingroup$ I think you're misreading the graph, which is why you're getting confused. The two curves are actually for the same element as a target material. What's different between the two curves is the acceleration voltage; the higher curve uses a larger accelerating voltage. The reason the larger accelerating voltage curve is above the low-voltage curve is because more energy is being imparted to the accelerated electrons in the high-voltage situation, and so the x-ray light emitted upon striking the target is more intense. $\endgroup$
    – DumpsterDoofus
    Commented May 17, 2014 at 20:01
  • $\begingroup$ I get that part, but let's assume for the moment we used the same accelerating voltage on two different elements. What would we expect to see in that case? $\endgroup$
    – Jesse
    Commented May 17, 2014 at 20:16
  • $\begingroup$ For the continuous bremsstrahlung see Kramer's Law. Intensity $I \propto Z$ (Atomic number). $\endgroup$
    – peanut_butter
    Commented May 17, 2014 at 22:03
  • $\begingroup$ Moseley found an empirical law relating the frequency of the characteristic lines to the atomic number of the target element. Quantum theory then provided an explanation of the law. en.wikipedia.org/wiki/Moseley%27s_law $\endgroup$
    – Farcher
    Commented May 8, 2016 at 19:49

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Both the intensity -wavelength curves are emission intensity of the same target material as the characteristic wavelengths are almost same. the minimum wavelength given by the two curves also indicates that the upper curve is due to higher tube voltage as wavelength (minimum) is proportional to (1/V) where V is the accelerating voltage .

In the referred book also -this fact has been mentioned that the higher curve has higher voltage.

The total x-ray energy emitted per second depends on the atomic number Z of the target material and on the x-ray tube current. This total x-ray intensity is given by
Intensity (cont) = A.i.Z. (V )^2 where A – proportionality constant i – tube current (measure of the number of electrons per second striking the target) V- tube voltage

Reference:http://web.stanford.edu/group/glam/xlab/MatSci162_172/LectureNotes/01_Properties%20&%20Safety.pdf

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